Functional Dependencies
Source:
The Theory of Relational Databases
D. Maier, Ed. Computer Science Press
Available at: http://www.dbis.informatik.hu-berlin.de/~freytag/Maier/
Cleveland State UniversityCIS 611 – Relational DatabasesPrepared by Victor Matos
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Functional Dependencies
• Two primary purposes of databases are to – attenuate data redundancy and – enhance data reliability.
• Any a priori knowledge of restrictions or constraints on permissible sets of data has considerable usefulness in reaching these goals.
• Data dependencies are one way to formulate such advance knowledge.
2
Example1
• Consider the relation
assign (Pilot, Flight, Date, Departs)PILOT FLIGHT DATE DEPARTS
CushingCushingClarkClarkClarkChinChinCopelyCopelyCopely
83116281301 83 83 116281281 412
9 AugIO Aug8 Aug12 Aug11 Aug13 Aug12 Aug9 Aug13 Aug15 Aug
10: 15a1:25p5:50a6:35p10: 15a10: 15a1:25p5:50a5:50a1:25p
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Example1 - Observations
• The relation assign tells which pilot flies a given flight on a given day, and what time the flight leaves.
• Not every combination of pilots, flights, dates, and times is allowable in assign. • The following restrictions apply, among others:
1. For each flight there is exactly one time.2. For any given pilot, date, and time, there is only one flight.3. For a given flight and date, there is only one pilot.
• These restrictions are examples of functional dependencies.
• Informally, a functional dependency occurs when the values of a tuple on one set of attributes uniquely determine the values on another set of attributes.
• Our restrictions can be phrased as
1. TIME functionally depends on FLIGHT,2. FLIGHT functionally depends on {PILOT, DATE, TIME}, and3. PILOT functionally depends on {FLIGHT, DATE}.
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FD DefinitionDef. Let r be a relation on scheme R, with X and Y subsets of R. Relation r satisfies the functional dependency (FD) X Y if for every X-value x,
y( X=x(r)) has at most one tuple.
One way to interpret this expression is to look at pairs of tuples, t1 and t2, in r.
If t1(X) = t2(X), then t1(Y) = t2(Y).
In the FD X Y the portion X is called the leftside and Y is called the right side.
5
FD Satisfies
Algorithm 4.1 SATISFIES
Input: A relation r and an FD X Y.
Output: true if T satisfies X Y, false otherwise.
SATISFIES(r, X Y);
1. Sort the relation r on its X columns to bring tuples with equal X-values together.
2. If each set of tuples with equal X-values has equal Y-values, return true. Otherwise, return false.
SATISFIES tests if a relation r satisfies an FD X Y.
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Algorithm: Satisfies
Using algorithm satisfies to test if FLIGHT DEPARTS
PILOT FLIGHT DATE DEPART
Cushing 83 9-Aug 10: 15a
Clark 83 11-Aug 10: 15a
Chin 83 13-Aug 10: 15a
Cushing 116 IO Aug 1:25p
Chin 116 12-Aug 1:25p
Clark 281 8-Aug 5:50a
Copely 281 9-Aug 5:50a
Copely 281 13-Aug 5:50a
Clark 301 12-Aug 6:35p
Copely 412 15-Aug 1:25p
Question:DEPARTS FLIGHT ???
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Inference Axioms
• The number of FDs that can apply to a relation r(R) is finite, since there is only a finite number of subsets of R.
• Thus it is always possible to find all the FDs that r satisfies, by trying all possibilities using the algorithm SATISFIES.
• This brute-force approach is time-consuming.
8
Inference Axioms
• Finding F requires semantic knowledge of the relation r.
• After knowing some members of F, it is often possible to infer other members of F.
• A set F of FDs implies the FD X Y, written
F X Y, if every relation that satisfies all the FDs in F also satisfies X Y.
• An inference axiom is a rule that states if a relation satisfies certain FDs, it must satisfy certain other FDs.
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Inference Axioms
F = { . . . }
Set of functional dependencies
Set of all relations r(R)
satisfying FDs in
F
X Y
A set F of FDs implies the FD X Y, written F X Y, if every relation that
satisfies all the FDs in F also satisfies X Y.
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Example - Inference Axioms
F = { A B, B C }Set of functional
dependencies
Set of all relations r(R)
satisfying FDs in
F
A C
A set F of FDs implies the FD X Y, written F X Y, if every relation that
satisfies all the FDs in F also satisfies X Y.
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Inference AxiomsThe Armstrong-Set of Inference Axioms • Axioms will implement the “intelligence” needed to
prove (or disprove) a sequence of derivations.
• Inference Machines are used to determine whether or not the application of the axioms on some ‘basic knowledge’ produces a ‘new’ valid piece of knowledge not there in the basic set.
• The first set we will consider is called the A-setproposed by W. Armstrong1.
1 William Armstrong: Dependency Structures of Data Base Relationships, page 580-583. IFIP Congress, 1974.
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A-AxiomsA1. Reflexivity X X
A2. Augmentation If (Z W; X Y) then XW YZ
A3. Additivity If { (X Y) and (X Z)} then X YZ
A4. Projectivity If (X YZ) then X Y
A5. Transitivity If (X Y) and (Y Z) then (X Z)
A6. Pseudotransitivity If (X Y) and (YZ W) then XZ W
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Inference Machine
INFERENCE MACHINE
A-AxiomsA1A2. . . A6
INPUT:Relation schema R
Set F of FDs on R
INPUT:
A “new” rule of the form X Y With X and Y in schema(R)
Output
YES
NO
Is the “new” rule XYderived from what is
known (R, F) by using the intelligence provided by the A-Axioms ?
If NO we must conclude that (F XY) is
not true
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Example1 - Using the A-AxiomsConsider R = (Street, Zip, City) ; and the dependencies
F = { City Street Zip, Zip City }
We want to show: Street Zip Street Zip City
Proof:
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Example1 - Using the A-AxiomsConsider R = (Street, Zip, City) ; and the dependencies
F = { City Street Zip, Zip City }
We want to show: Street Zip Street Zip City
Proof:
1. Zip City – Given
2. Street Zip Street City – Augmentation of (1) by Street
3. City Street Zip – Given
4. City Street City Street Zip – Augmentation of (3) by
City Street
5. Street Zip City Street Zip – Transitivity of (2) and (4)
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Example2 – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies
F = { ABE AGJ BE I E G GI H }
Show that
AB GH
is derived by F.
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If YES give a proof
If NO provide a counter-example
Example2A – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies
F = { ABE AGJ BE I E G GI H }
Show that
AB GH
is derived by F.
Step Statement Explanation
1 AB E Given
2 E G Given
3 AB G Transitivity on (1) and (2)
4 AB BE Augmentation (1) by B
5 BE I Given
6 AB I Transitivity on (4) and (5)
7 AB GI Additivity on (6) and (3)
8 GI H Given
9 AB H Transitivity on (7) and (8)
10 AB GH Additivity on (3) and (11)Q.E.D.
quod eratdemonstrandum
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Example2B – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies
F = { ABE AGJ BE I E G GI H }
Show that
AB GH
is derived by F.
Step Statement Explanation
1 AB E Given
2 AB AB Reflexivity
3 AB B Projectivity on (2)
4 AB BE Additivity on (1) and (3)
5 BE I Given
6 AB I Transitivity on (4) and (5)
7 E G Given
8 AB G Transitivity on (1) and (7)
9 AB GI Additivity on (6) and (8)
10 GI H Given
11 AB H Transitivity on (9) and (10)
12 AB GH Additivity on (8) and (11)Q.E.D.quod eratdemonstrandum
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again!
Example3 – Using A-AxiomsConsider the relation schema <R,F> where R = (ABCDEGHI) and dependencies
F = { ABE AGJ BE I E G GI H }
Show that
AEI H
is derived by F.
Step Statement Explanation
1
2
3
4
5
6
7
8
9
10
11
12 20
Your turn!
Reducing the A-Axioms
The set of A-Axioms is not minimal, therefore some of its rules could be eliminated.
Observations
• Rule A5 (transitivity) is a special case of rule A6 (pseudo-transitivity).
• Rules A3 (additivity) and A4 (projectivity) can be derived from A1 (reflexivity), A2 (augmentation), A6 (pseudo-transitivity).
Proof
(a) First observation is trivial (just make Z= Ø)
(b) Axiom A3 (Additivity) states that two rules, say X Y and X Z, can be combined in one X YZ. Lets use A2 on X Y to produce XZ YZ. Repeat A2 this time on X Z to produce X XZ. Now apply A5 on X XZ and XZ YZ; we get X YZ. Therefore, we conclude that X YZ without using the rule A3 itself (see next page)
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Reducing the A-Axioms
The set of A-Axioms is not minimal, therefore some of its rules could be eliminated.
Statement
Axiom A3 is redundant. Rule A3 (Additivity) states that two rules, say X Y and X Z, can be combined in one X YZ.
Proof
We can prove that this fact is true without using A31. X Y Given
2. XZ YZ (A2) Augmenting (1) by Z
3. X Z Given
4. X XZ (A2) Augmenting (3) by Z
5. X YZ (A5) Transitivity on (4) and (2)
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Characterizing the A-Axioms
• The set of A-Axioms is complete
Therefore every FD that is implied by a set F of FDs can be derived from the FDs in F and one or more applications of the A-Axioms ( FA XY )
• A-Axioms are correct
Applying the axioms to FDs in a set F can only produce FDs that are implied by F.
• The set of A-axioms is not minimal
Some rules are added for convenience but they can be removed without diminishing the expressive power of the A-axioms
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Correctness of the A-Axioms
The axioms can not be used to prove a false derivation. In such a case showing a counter-example is sufficient to establish the falsity of a statement.
Example
Assume schema R(XYZW). Does ( XY ZW ) A X Z ?
The correct answer is NO. To show support for our argument we produce a counter-example. For instance:
On the example table there are no violations to the fact that XY implies a unique ZW (12 34 and 15 67). However X=1 determines two different Z values, 3 and 6. Therefore X Z is not a valid dependency as shown in the counter-example.
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X Y Z W
1 2 3 4
1 5 6 7
F+
Closure F+
• Let F be a set of FDs for a relation r(R). The closure of F, denoted F+, is the smallest set containing F such that the A-axioms cannot be applied to the set to produce a new rule not included in the set already
• Since F+ must be finite, we can compute it by starting with F, applying A1, A2, and A6, and adding the derived FDs to F until no new FDs can be derived.
• The closure of F depends
on the scheme R.
• If R = (A B) then F+ will always
contain B B, but if R = (A C),
F+ never contains B B.
F
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Closure F+
• The set F derives an FD X Y if X Y is in F+.
• Since our inference axioms are correct, if F derives X Y, then F implies X Y ( F A X Y )
• Note that F+ = (F+)+
• It is desirable to determine whether F A X Y without computing F+
• Computing the entire set F+ is time-consuming and tedious
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Closure F+
Example: Consider the relation schema <R,F>
where R = (A B C) and F = { AB C, C B }.
By the use of brute-force we produce all rules out of F.
F+ is the set of rules listed below
F+
F = { ABC, CB }
A A
B B
C C
C B
AB AB
AC AC
BC BC
AB C
AB A
AB B
…
BC C
ABC ABC
ABC A
ABC B
ABC C
ABC AB
…
ABC BC
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A A
B B
C C
C B
AB AB
AC AC
BC BC
AB C
AB A
AB B
…
BC C
ABC ABC
ABC A
ABC B
ABC C
ABC AB
…
ABC BC
Closure F+
Example: Consider the relation schema <R,F> where R = (ABC)
and F = { AB C, C B }.
Question: Does F B C ?
Answer: F+ is the set of rules listed below and B C is not in the set; therefore the rule B C is not implied by F.
F+
F = { ABC, CB }
B C
This rule is NOT reachable from F
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Closure F+
Aside: How many FDs are there in <R,F>
An upper bound is
Each sum term represents the possible combinations of r attributes made out of the total n domains for each of the m X Y rule in F.
for n=3 there are (23-1)2 = 49 possibilities, however for R
holding 10 attributes there are over a million possible FDs
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2
1 1
(2 1)n n
n
r r
n n
r r
Closure F+
Definition. An FD X Y is trivial if X Y.
• If F is a set of FDs over R and X is a subset of R, then there is a FD X Y in F+ such that Y is maximal: for any other FD X Z in F+, Y Z.
– This result follows from additivity. – The right side Y is called the closure of X and is
denoted by X+.
• The closure of X always contains X, by reflexivity.
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Derivations and DDAGs
• If F XY, then either XY is in F, or a series of applications of the inference A-axioms to F will yield X Y.
• This sequence of axiom applications and resulting FDs is called a derivation of X Y from F.
• More formally, let F be a set of FDs over scheme R . A sequence P of FDs over R is a derivation sequence on F if every FD in P either– is a member of F, or– follows from previous FDs in P by an application of one of the inference
axioms A1 to A6.
• P is a derivation sequence for XY if XY is one of the FDs in P.
• DefinitionLet P be a derivation sequence on F. The use set of P is the collection of all FDs (originally) in F that appear in P.
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Derivations and DDAGsEXAMPLE
Consider schema r(ABCDEG) and functional dependencies F = { A BC, BD G, C ED }
A derivation sequence for A E is
Try…
Step Explanation
12345
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Derivations and DDAGsEXAMPLE
Consider schema r(ABCDEG) and functional dependencies F = { A BC, BD G, C ED }
A derivation sequence for A E is
The set P for AE (five rules written above) is a derivation sequence on F. The Use_Set_Of_P is = {A BC, C ED }
Step Explanation
12345
A BCA C C EDC EA E
(given)(Projectivity [A4] on 1)(given)(Projectivity[A4] on 3)(Transitivity[A5] on 2 and 4)
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Derivations and DDAGsExample. Consider schema <R, F> where R= { A B C D E G H I J }
and F = { ABE, AGJ, BE I, E G, GI H}The following sequence is a derivation sequence for A B G H.
This P sequence contains unneeded FDs, such as 12 and 13, and is also a derivation sequence for other FDs, such as A B G I.The Use_Set_Of_P is {C ED, BE I, E G, GI H }
Step Explanation
1. AB E2. AB AB3. AB B4. AB BE5. BE I6. AB I7. E G8. AB G9. AB GI
10. GI H11. AB H12. GI GI13. G I I14. ABG H
given)(reflexivity)(projectivity from 2)(additivity from 1 and 3)(given)(transitivity from 4 and 5)(given)(transitivity from 1 and 7)(additivity from 6 and 8)(given)(transitivity from 9 and 10)(reflexivity)(projectivity from 12)(additivity from 8 and 11)
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B-AxiomsDefinition:
The B-Axioms set is a small and complete collection of inference rules. It is not a subset of A1 to A6, however it is equally expressive. For a relation (R), with W, X, Y, and Z subsets of R, and C an attribute in R then:
B1. Reflexivity X X
B2. Accumulation If (X YZ) and (Z CW) then X YZC
B3. Projectivity If (X YZ) then X Y
Motivation:This is another approach to the problem of finding a sequence of derivations using a smaller set of axioms.
Significance:Since B-Axioms are complete, we can always find a derivation sequence using only the three B-axioms to assert whether or not FB XY.
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B-AxiomsExample: Let R = (ABCDEGHI) and F = {ABE
AG J BE I E G GI H}
ProblemFind a derivation sequence P showing FB AB GH using only B-axioms.
AnswerSee P sequence on the right
Comment(a) Use_Set_Of_P contains
rules on lines 2, 5, 9, 11.(b) Too many steps!
Step Explanation
1
2
3
4
5
6
7
8
9
10
11
12
13
14
36
B-AxiomsExample: Let R = (ABCDEGHI) and F = {ABE
AG J BE I E G GI H}
ProblemFind a derivation sequence P showing FB AB GH using only B-axioms.
AnswerSee P sequence on the right
Comment(a) Use_Set_Of_P contains
rules on lines 2, 5, 9, 11.(b) Too many steps!
Step Explanation
1 EI EI Reflexivity (B1)
2 E G Given
3 EI EIG Accumulation (B2)
4 EI GI Projectivity (B3) from (3)
5 GI H Given
6 EI GHI Accumulation from (4) and (5)
7 EI GH Projectivity from (6)
8 AB AB Reflexivity
9 AB E Given
10 AB ABE Accumulation from (8) and (9)
11 BE I Given
12 AB ABEI Accumulation from (10) and (11)
13 AB ABEIG Accumulation from (4) and (12)
14 AB ABEGHI Accumulation from (7) and (13)
15 AB GH Projectivity from (14) 37
Ok, but useless
RAP-Derivation SequenceRAP: Stands for: Reflexivity, Augmentation, Projectivity
Definition: Consider derivation sequences for X Y on a set F of FDs using the B-axioms that satisfy the following constraints:1. The first FD is X X2. The last FD is X Y3. Every FD other than the first and last is either an FD in F (given) or
and FD of the form X Z that was derived using axiom B2 (Accumulation).
Such a derivation is called a RAP-derivation sequence
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RAP-Derivation SequenceExample:
Let R = (ABCDEGHI) and
F = { ABE AGJ BE I E G GI H }
Find a RAP-sequence for AB GH
Comments
1. The table contains a RAP sequence
for ABGH.
2. Each rule in P is either given in F
or the result of applying B2 on
previous rules in P.
3. First and Last lines agree with the
definition of RAP sequence.
4. Use_Set_Of_P contains rules in
lines 2, 4, 6, 8.
Step Explanation1 AB AB B1
2 AB E Given
3 AB ABE B2
4 BE I Given
5 AB ABEI B2
6 E G Given
7 AB ABEIG B2
8 GI H Given
9 AB ABIGH B2
10 AB GH B3
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RAP-Derivation SequenceExample:
Let R = (ABCDEGHI) and
F = { ABE AGJ BE I E G GI H }
Find a RAP-sequence for BHE GI
Your turn…
Step Explanation1
2
3
4
5
6
7
8
9
10
40
RAP-Derivation SequenceExample:
Let R = (ABCDEG) and
F = { A BC, BD G, C ED }
Find a RAP-sequence for AD GE
Your turn…
Step Explanation1
2
3
4
5
6
7
8
9
10
41
RAP-Derivation SequenceExample:
Let R = (ABCDEI) and
F = { A D, AB E, BI E, CD I, E C}
Find a RAP-sequence for AE DCI
Your turn…
Step Explanation1
2
3
4
5
6
7
8
9
10
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Derivation DAGsA directed acyclic graph (DAG) is a directed graph with no directed paths from any node to itself.
A labeled DAG is a DAG with an element from some labeling set L associated with each node.
Valid DAG (disconnected but OK)
Not a Valid DAG (Path: A-D-A makes a cycle)
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Derivation DAGs• DAGS are a convenient way of graphically representing a
derivation sequence of the form F B X Y
• Whenever there is a RAP derivation sequence there is an equivalent DDAG (and conversely)
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Derivation DAGsEXAMPLE
Consider schema r(ABCDEG) and functional dependencies
F = { A BC, BD G, C ED }. Show a DDAG for AD GE
NOTE:
The Use_Set of the derivation sequence is { A BC, BD G, C E }
A
D
B
C
G
E
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Derivation DAGsRules for Constructing a DDAG
Rule 1. Any set of unconnected nodes with labels from r(R) is an F-based DDAG
Rule 2. Let H be a DDAG including nodes labeled A1 … Ak. Let rule A1…Ak B be part of F. Form graph H’ by adding a new node labeled “B” and new edges <A1,B>,…,<AK,B>.
Rule 3. Nothing else is an F-based derivation DDAG.
A1 A2
…
A1
AK
AN
B
New edges
New node
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Derivation DAGsExample
Consider the relation schema r(ABCDEGHIJ) subject to the dependencies
in F = { AB E, AG J, BE I, E G, GI H }.
Draw a DDAG for rule AB GH
Note: The Use_Set of the derivation sequence is { AB E, BE I, E G, GI H }
A
B
E
I
G
H
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Derivation DAGsExample
Consider the relation schema r(ABCDEGHIJ) subject to the dependencies
in F = { AB E, AG J, BE I, E G, GI H }.
Draw a DDAG for the new rule BIG JA
NOTE: No path from source to destination is possible, therefore the new rule BIG AJ is not derivable from F.
B
I H A
G
J
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Derivation DAGsExample
Consider the relation schema R(ABCDEGHIJ) subject to the dependencies
in F = { AB E, AG J, BE I, E G, GI H }.
Draw a derivation DDAG for the new rule AB HC
NOTE: Node C is not reachable from the source. Therefore the rule AB CH cannot be deduced from F.
A
B
E
I
C
H
G
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X + Closure of a Set of Attributes
• In order to simplify the asserting of whether or not a rule X Y follows from a set F of FDs, we will compute X+ the closure of a set of attributes X
• The set X+ is the maximal set of attributes which can be derived from X using a RAP derivation sequence starting on X
• We will say that X Y is in F+ whenever Y is in X+
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Computing X +
The following algorithm to compute X+ has poor performance but is easy to understand
Algorithm: X-ClosureInput: A set of attributes X and a set of FDs FOutput: The closure of X under F denoted X+
function X-CLOSURE (X, F)begin
OldDep = ; NewDep = X;while ( NewDep OldDep ) do begin
OldDep = NewDepfor every FD A B in F do
if ( NewDep A ) then NewDep = NewDep B;end while;return ( NewDep )
end function;
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Computing X +
EXAMPLES
Consider the relation schema r(ABCDEGHIJ) subject to the dependencies in
F = { AB E, AG J, BE I, E G, GI H }.
Compute closure of AB
(AB) + = A B reflexivity
A B E using AB E
A B E I BE I
A B E I G E G
A B E I G HJ GI H
nothing else could be added to AB+
Note: Observe that AB ABEIHG. This rule is a compact notation for the 27 FDs having AB as LHS.52
Computing X +
EXAMPLES
Consider the relation schema r(ABCDEGHIJ) subject to the dependencies in
F = { AB E, AG J, BE I, E G, GI H }.
Compute closure DEC
(DEC) + = D E C
D E C G using E G
nothing else could be added
53
Member Algorithm
Checking Membership
In order to verify whether or not a functional dependency X Y could be derived from a set F of FDs the following simple test could be applied
F B X Y if Y is part of X+
54
Member AlgorithmMember Algorithm
Input: Rule X Y and functional dependencies F
Output: TRUE whenever the rule is derived from F
Method:
begin
if ( Xclosure (X, F) Y ) then
return( True )
else
return( False );
end;
Example
Question: Does rule AB EH follow from F = { AB E, AG J, BE I, E G, GI H }
Answer: YES. Observe that (AB)+ = ABEIGHJ EH.
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Linear Closure – XF+
Input: A set of attributes X and a set of functional dependencies F
Output: The closure of X under F demote XF+
Procedure LINCLOSURE ( Attribute X, SetOfFDs F)
BEGIN
/* Initialization */
for each FD W Z in F do begin
COUNT[ W Z ] = lenghtOf(w);
for each attribute A in W do
add rule W Z into LIST[ A ];
end;
NEWDEP = X; UPDATE = X;
/* Computation */
while ( UPDATE Ø) do begin
Choose an attribute A in UPDATE;
UPDATE = UPDATE - A;
for each FD W Z in LIST[A] do begin
COUNT[W Z] = COUNT[W Z] - 1;
if ( COUNT[w Z] = 0 ) then
ADD = Z - NEWDEP;
NEWDEP = NEWDEP ADD;
UPDATE = UPDATE ADD;
end if;
end for;
end while;
END 56
Linear Closure – XF+
Example
Consider the schema r(ABCDEI) subject to the dependencies
F = { A D, AB E, BI E, CD I, E C}
Find the closure of AE using the Linear Closure algorithm
57
Linear Closure – XF+
Example (continuation…) Tracing the execution of the linear time closure algorithm
58
UPDATE= AE NEWDEP= AE
List[A] Rule 1 A D Count[1] = 0therefore add D to both stringsRule 2 AB E Count[2] = 1
UPDATE= ED NEWDEP= AED
List[E] Rule 5 E C Count[5] = 0therefore add C to both strings
UPDATE= DC NEWDEP= AEDC
List[D] Rule 4 CD I Count[5] = 1
UPDATE= CNEWDEP= AEDC
List[C] Rule 4 CD I Count[4] = 0therefore add I to both strings
UPDATE= INEWDEP= AEDCI
List[I] Rule 3 BI E Count[3] = 1
UPDATE= ØNEWDEP= AEDCI
therefore (AE) + = AEDCI
Linear Closure – XF+
Homework
You will create a CASE tool for designing ‘good’ databases.
The first step involves the implementation of the XLinearClosure() algorithm.
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